Proving Convergence of a Series

  • Thread starter hy23
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  • #1
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Homework Statement


([tex]\sqrt{(n+1)}[/tex] - [tex]\sqrt{n}[/tex] ) [tex]/[/tex] [tex]\sqrt{n}[/tex]
I'm trying to show this series converges.


Homework Equations


divergence test: as n approaches infinity, if the sequence does not approach 0 then the series diverges
ratio test: as n approaches infinity, if the ratio between subsequent terms is less than 1 then the series converges, if equal to 1 the test is inconclusive, if greater than 1 then the series diverges
comparison test: if this series is smaller than another series that converges, then this series also converges; if this series is greater than another series that diverges, then this series also diverges



The Attempt at a Solution


Applying the divergence test, the sequence definitely approaches 0, but this cannot be used to prove convergence
I tried using ratio test (which has worked so well for nearly every question) and ended up with a mess of square roots.
I'm thinking of using comparison test, but what series can I compare it to?
 

Answers and Replies

  • #2
hunt_mat
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Note that:
[tex]
1=n+1-n=(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})
[/tex]
and hence
[tex]
\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}
[/tex]
 
  • #3
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Sorry but I don't see where you're going with this, can you explain?
 
  • #4
He wants you to rationalise the numerator.
 
  • #5
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I know what he did there but I don't know how that step leads to proving convergence
 
  • #6
hunt_mat
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Ideally I would like you to compare your series to the harmonic series which is known to diverge
 
  • #7
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yes hunt, but this series is smaller than the harmonic series, thus you can't compare it to the harmonic to prove divergence, maybe it's more plausible to compare it to 1/2n, and use the limit comparison test instead?
 
  • #8
yes hunt, but this series is smaller than the harmonic series, thus you can't compare it to the harmonic to prove divergence, maybe it's more plausible to compare it to 1/2n, and use the limit comparison test instead?

[tex] 2 \sqrt{n+1} > \sqrt{n} + \sqrt{n+1}[/tex]

[tex]\frac{1}{ 2 \sqrt{n+1}} < \frac{1}{ \sqrt{n} + \sqrt{n+1}}[/tex]

[tex] \frac{1}{2(n+1)} < \frac{1}{ 2 \sqrt{n+1}}[/tex]

The series diverges.
 

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